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Ditanya pers garis tegak lurus?
Dijawab :
3x = 5y + 1
3x - 5y - 1 =0
m1 = -a / b = -(3/-5) = 3/5
Syarat tegak lurus m1.m2=-1
m1.m2=-1
3/5.m2=-1
m2 = -5/3
Pers garis melalui titik M(-2,5) dan m2=-5/3
y - y1 = m (x - x1)
y - 5 = -5/3 (x + 2)
3y - 15 = -5x -10
5x + 3y - 5 =0
3x = 5y + 1
5y = 3x - 1
y = 3x/5 - 1/5
m1 = 3/5
tegak lurus :
m1 × m2 = -1
3/5 × m2 = -1
m2 = -1/ 3/5 = -1 × 5/3 = -5/3
persamaan garis :
y - y1 = m(x - x1)
y - 5 = -5/3 (x - (-2))
[y - 5 = -5/3 (x + 2)] × 3
3y - 15 = -5x - 10
5x + 3y - 15 + 10 = 0
5x + 3y - 5 = 0